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A book has 1000 pages numbered 1, 2, 3, …, and so on. How

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by Max@Math Revolution » Mon May 06, 2019 11:57 pm
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[GMAT math practice question]

A book has 1000 pages numbered 1, 2, 3, …, and so on. How many times does the digit 2 appear on the page numbers?

A. 200
B. 250
C. 300
D. 400
E. 500
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by GMATGuruNY » Tue May 07, 2019 2:32 am
Max@Math Revolution wrote:[GMAT math practice question]

A book has 1000 pages numbered 1, 2, 3, …, and so on. How many times does the digit 2 appear on the page numbers?

A. 200
B. 250
C. 300
D. 400
E. 500
Ignore page 1000, since it does not include the digit 2.
To make the calculation easier, consider the remaining pages numbered as 3-digit integers, beginning with 000 and ending with 999:
000, 001, 002...997, 998, 999

There are 1000 options between 000 to 999, inclusive.
Since each option is composed of 3 digits, the total number of digits = 3*1000 = 3000.
The probability that 2 will appear in any given position is the same as the probability that 3 will appear in any given position.
Implication:
Each of the 10 digits 0 through 9 will appear the SAME NUMBER OF TIMES.
Thus, the number of times that 2 will appear = 3000/10 = 300.

The correct answer is C.
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by Max@Math Revolution » Thu May 09, 2019 1:33 am
=>

The page numbers 2, 12, 20, 21, 22, … , 29, 32, 42, … , 92 contain 19+1 = 20 copies of the digit, 2.
The page numbers 102, 112, 120, 121, … , 129, 132, 142, … , 192 similarly contain 20 copies of the digit 2.
The page numbers 200, 201, … , 299 contain 120 copies of the digit, 2.
Page numbers 302, 312, 320, 321, … , 329, 332, 342, … , 392 contain 20 copies of the digit, 2.
…
Page numbers 902, 912, 920, 921, 922, … , 929, 932, 942, … , 992 contain 20 copies of the digit, 2.

The total number of copies of the digit, 2 printed on the pages is
9*20 + 120 = 300.

Therefore, the answer is C.
Answer: C
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